Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Some bifurcation results for a semilinear elliptic equation Francesca Gladiali University of Sassari, Italy, fgladiali@uniss.it Optimization Days Ancona, June 6-8, 2011 joint work with M.Grossi, F.Pacella and P.N.Srikanth.
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Our problem We consider the problem − ∆ u = u p in Ω u > 0 in Ω (1) u = 0 on ∂ Ω R N : a < | x | < b } , b > a > 0, is an where either Ω = A := { x ∈ I R N \ B 1 (0), is the exterior annulus, N ≥ 2, p ∈ (1 , + ∞ ), or Ω = I of a ball, N ≥ 3 and p > N +2 N − 2 .
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The case of the annulus We consider first the case of the annulus Ω = A . Problem (1) has a radial solution u = u ( A , p ) [Kazdan-Warner (1975)], and this radial solution is unique [Ni-Nussbaum (1985)].
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The case of the annulus We consider first the case of the annulus Ω = A . Problem (1) has a radial solution u = u ( A , p ) [Kazdan-Warner (1975)], and this radial solution is unique [Ni-Nussbaum (1985)]. We study the structure of the set of nonradial solutions which bifurcate from the radial solutions of (1) varying the domain A or the exponent p .
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The case of the annulus We consider first the case of the annulus Ω = A . Problem (1) has a radial solution u = u ( A , p ) [Kazdan-Warner (1975)], and this radial solution is unique [Ni-Nussbaum (1985)]. We study the structure of the set of nonradial solutions which bifurcate from the radial solutions of (1) varying the domain A or the exponent p . The first step in studying the bifurcation is to analyze the possible degeneracy of the radial solution u depending on the annulus or on the exponent, i.e. see if the linearized operator L u := − ∆ − pu p − 1 I admits zero as an eigenvalue.
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Radial Nondegeneracy The Linearized Problem is � − ∆ v − pu p − 1 v = 0 in A , (2) v = 0 on ∂ A and we want to see whether solutions do exist.
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Radial Nondegeneracy The Linearized Problem is � − ∆ v − pu p − 1 v = 0 in A , (2) v = 0 on ∂ A and we want to see whether solutions do exist. Lemma The linearized problem does not admit any nontrivial radial solution. The radial Morse index of u is 1.
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem It is easy to see that solving L u ( v ) = 0, i.e. � − ∆ v − pu p − 1 v = 0 in A , v = 0 on ∂ A , is equivalent to show that the linear operator L u := | x | 2 � � � − ∆ − pu p − 1 I , x ∈ A (3) has zero as an eigenvalue with the same boundary conditions.
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem It is easy to see that solving L u ( v ) = 0, i.e. � − ∆ v − pu p − 1 v = 0 in A , v = 0 on ∂ A , is equivalent to show that the linear operator L u := | x | 2 � � � − ∆ − pu p − 1 I , x ∈ A (3) has zero as an eigenvalue with the same boundary conditions. Consider the 1-dimensional operator � � − v ′′ − N − 1 v ′ − pu p − 1 v � L u ( v ) := r 2 r ∈ ( a , b ) (4) r with the same boundary conditions.
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem The spectra of these operators are related by σ ( � L u ) = σ ( � L u ) + σ ( − ∆ S N − 1 ) where − ∆ S N − 1 is the Laplace-Beltrami operator on the sphere S N − 1 .
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem The spectra of these operators are related by σ ( � L u ) = σ ( � L u ) + σ ( − ∆ S N − 1 ) where − ∆ S N − 1 is the Laplace-Beltrami operator on the sphere S N − 1 . Let us denote by α j = α j ( A , p ) the eigenvalues of � L u and by λ k = k ( k + N − 2) the eigenvalues of − ∆ S N − 1 , the question is whether there exists j and k such that 0 = α j ( A , p ) + λ k .
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem Theorem The linearized equation L u ( v ) = 0 (2) has a nontrivial solution ψ ( x ) if and only if α 1 ( A , p ) + λ k = 0 (5) for some k ≥ 1 . Moreover these solutions have the form ψ ( x ) = w 1 ( | x | ) φ k ( x | x | ) . Here α 1 and w 1 are the first eigenvalue and the first eigenfunction of the radial operator � L u and φ k is an eigenfunction of the Laplace-Beltrami operator relative to the eigenvalue λ k .
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem - Only the first eigenvalue of the operator � L u is responsible for degeneracy of the radial solutions.
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem - Only the first eigenvalue of the operator � L u is responsible for degeneracy of the radial solutions. - Only the eigenvalue α 1 ( A , p ) depends on the annulus A and on the exponent p , while λ k depends only on the dimension N . Indeed λ k = k ( k + N − 2).
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem - Only the first eigenvalue of the operator � L u is responsible for degeneracy of the radial solutions. - Only the eigenvalue α 1 ( A , p ) depends on the annulus A and on the exponent p , while λ k depends only on the dimension N . Indeed λ k = k ( k + N − 2). So in order to study the equation α 1 ( A , p ) + λ k = 0 we have to analyze the dependence of α 1 on A and p .
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems The Linearized Problem - Only the first eigenvalue of the operator � L u is responsible for degeneracy of the radial solutions. - Only the eigenvalue α 1 ( A , p ) depends on the annulus A and on the exponent p , while λ k depends only on the dimension N . Indeed λ k = k ( k + N − 2). So in order to study the equation α 1 ( A , p ) + λ k = 0 we have to analyze the dependence of α 1 on A and p . Recent results by [T.Bartsch-M.Clapp-M.Grossi-F.Pacella(2010), F.G.-M.Grossi-F.Pacella-P.N.Srikanth(2010)]
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Varying the exponent p R N : a < | x | < b } , and let the Now we fix the annulus A = { x ∈ I exponent p vary. So we write u = u p and α 1 = α 1 ( p ).
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Varying the exponent p R N : a < | x | < b } , and let the Now we fix the annulus A = { x ∈ I exponent p vary. So we write u = u p and α 1 = α 1 ( p ). The solution u p admits a limiting problem as p �→ + ∞ . Theorem (M.Grossi(2006)) Let u p be the unique radial solution of (1). Then as p → + ∞ 4( N − 2) in C 0 (¯ u p ( | x | ) → a 2 − N − b 2 − N G ( r , r 0 ) A ) and also in H 1 0 , r ( A ) , where r 0 ∈ ( a , b ) and G ( r , s ) is the Green’s function of the operator − ( r N − 1 u ′ ) ′ , r ∈ ( a , b ) with Dirichlet boundary conditions. Moreover � u p � ∞ = 1 + log p p + o (1 + γ p ) , γ > 0 , as p → + ∞ . p
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Varying the exponent p Moreover Theorem (M.Grossi(2006)) Letting p u p ( r ) = ˜ ( u p ( ǫ p r + r p ) − � u p � ∞ ) , � u p � ∞ p � u p � p − 1 = 1 , we have that where u p ( r p ) = � u p � ∞ and p ǫ 2 in C 1 ˜ u p → U loc ( I R ) , (6) where U is the unique solution of � − U ′′ = e U in I R U ′ (0) = 0 U (0) = 0
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Varying the exponent p For the first eigenvalue α 1 ( p ) we get: the map p �→ α 1 ( p ) is real analytic;
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Varying the exponent p For the first eigenvalue α 1 ( p ) we get: the map p �→ α 1 ( p ) is real analytic; 0 p 2 + o ( p 2 ) (so α 1 ( p ) → −∞ as p → + ∞ ); α 1 ( p ) = − 1 2 γ r 2
Presentation of the problem The case of the annulus The case of the exterior of a ball Some open problems Varying the exponent p For the first eigenvalue α 1 ( p ) we get: the map p �→ α 1 ( p ) is real analytic; 0 p 2 + o ( p 2 ) (so α 1 ( p ) → −∞ as p → + ∞ ); α 1 ( p ) = − 1 2 γ r 2 α 1 ( p ) → 0 as p → 1 .
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